等待具有多个并行作业的子进程结束

2024-06-25 06:41:15 发布

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我正在并行运行python中的一些子进程。我想等到每个子进程都完成。我正在做一个非优雅的解决方案:

runcodes = ["script1.C", "script2.C"]
ps = []
for script in runcodes:
  args = ["root", "-l", "-q", script]
  p = subprocess.Popen(args)
  ps.append(p)
while True:
  ps_status = [p.poll() for p in ps]
  if all([x is not None for x in ps_status]):
    break

是否有一个类可以处理多个子进程?问题是wait方法阻塞了我的程序。

更新:我想显示计算过程中的进度:类似于“4/7子进程完成…”

如果您有兴趣,请编译c++脚本并执行它。


Tags: infor进程statusscriptargsroot解决方案
3条回答

怎么样

import os, subprocess
runcodes = ["script1.C", "script2.C"]
ps = {}
for script in runcodes:
    args = ["root", "-l", "-q", script]
    p = subprocess.Popen(args)
    ps[p.pid] = p
print "Waiting for %d processes..." % len(ps)
while ps:
    pid, status = os.wait()
    if pid in ps:
        del ps[pid]
        print "Waiting for %d processes..." % len(ps)

你可以这样做:

runcodes = ["script1.C", "script2.C"]

ps = []
for script in runcodes:
    args = ["root", "-l", "-q", script]
    p = subprocess.Popen(args)
    ps.append(p)

for p in ps:
    p.wait()

进程将并行运行,您将在最后等待所有进程。

如果您的平台不是Windows,那么您可以选择子进程的stdout管道。你的应用程序将被阻止,直到:

  • 其中一个注册的文件描述符有一个I/O事件(在本例中,我们对子进程的stdout管道上的挂断感兴趣)
  • 投票超时了

在Linux 2.6.xx中使用epoll的非充实示例:

import subprocess
import select

poller = select.epoll()
subprocs = {} #map stdout pipe's file descriptor to the Popen object

#spawn some processes
for i in xrange(5):
    subproc = subprocess.Popen(["mylongrunningproc"], stdout=subprocess.PIPE)
    subprocs[subproc.stdout.fileno()] = subproc
    poller.register(subproc.stdout, select.EPOLLHUP)

#loop that polls until completion
while True:
    for fd, flags in poller.poll(timeout=1): #never more than a second without a UI update
        done_proc = subprocs[fd]
        poller.unregister(fd)
        print "this proc is done! blah blah blah"
        ...  #do whatever
    #print a reassuring spinning progress widget
    ...
    #don't forget to break when all are done

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